The eccentric angles of the extremities of latus rectum of the ellipse a 2 x 2 + b 2 y 2 =1 are given by:
Question
The eccentric angles of the extremities of latus rectum of the ellipse are given by:
Solution
The latus rectum of an ellipse is the line segment that passes through the focus of the ellipse, is perpendicular to the major axis and both its endpoints lie on the ellipse.
The equation of the ellipse is given by a^2x^2 + b^2y^2 = 1.
The eccentricity of the ellipse is given by e = sqrt(1 - (b^2/a^2)) if a>b (ellipse is horizontal) or e = sqrt(1 - (a^2/b^2)) if b>a (ellipse is vertical).
The extremities of the latus rectum for the ellipse are given by (ae, b^2/a) and (ae, -b^2/a) for a horizontal ellipse and (a^2/b, be) and (-a^2/b, be) for a vertical ellipse.
The eccentric angles of these points are given by the inverse sine of the y-coordinate divided by b for a horizontal ellipse and the inverse sine of the x-coordinate divided by a for a vertical ellipse.
So, the eccentric angles of the extremities of the latus rectum of the ellipse are given by sin^-1(b^2/(ae)) and -sin^-1(b^2/(ae)) for a horizontal ellipse and sin^-1(a^2/(be)) and -sin^-1(a^2/(be)) for a vertical ellipse.
Similar Questions
Find the eccentricity of an ellipse whose length of the minor axis is equal to half of the length between foci.
S and T are the foci of an ellipse and B is an end of the minor axis. If STB is an equilateral triangle, the eccentricity of the ellipse is
The equation(s) of the tangent(s) to the ellipse 9(x - 1)2 + 4y2 = 36 parallel to the latus rectum, is (are)
Find the area of the largest rectangle that can be inscribed in the ellipse x2a2 + y2b2 = 1.
What are the foci of the ellipse" (x-4)^2/ 4,+ (y+1)^2 / 9+ =1?(4,-1√5)(4,-1√5)(4, −1 ± √3)(3, −1 ± 3)(2,-1+√5)
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.